Second Law, Entropy & Availability

The second law is a signature GATE ME topic (~2 marks). Carnot efficiency, COP, entropy, and availability are heavily and repeatedly tested.

Second law & Carnot

Statements: Kelvin–Planck (no engine can convert all heat to work in a cycle) and Clausius (heat cannot flow cold → hot without work). Consequence — the Carnot cycle sets the maximum efficiency between two reservoirs:

• Heat-engine efficiency η = W/Q_h = 1 − Q_c/Q_h; Carnot η = 1 − T_c/T_h (T in kelvin).
• Refrigerator COP = Q_c/W = T_c/(T_h − T_c); heat-pump COP = Q_h/W = T_h/(T_h − T_c). Note COP_HP = COP_ref + 1.
  Real cycles fall below Carnot due to irreversibility.

Entropy & availability

Clausius inequality: ∮δQ/T ≤ 0 (= 0 reversible, < 0 irreversible). Entropy dS = (δQ/T)_rev; entropy is a property. For an isolated system, ΔS ≥ 0 (increase-of-entropy principle). Entropy generation S_gen ≥ 0 measures irreversibility.

Availability (exergy) = maximum useful work obtainable as a system comes to equilibrium with the surroundings (dead state, T₀). Availability is destroyed by irreversibility: work lost = T₀·S_gen (Gouy–Stodola).

Exam Tricks & Tips

• 🎯 Carnot η = 1 − T_c/T_h with temperatures in KELVIN — the single most-tested second-law relation.
• 🎯 COP_ref = T_c/(T_h−T_c), COP_HP = T_h/(T_h−T_c), and COP_HP = COP_ref + 1 — three linked one-liners.
• 🎯 ΔS ≥ 0 for an isolated system; S_gen = 0 only for a reversible process — the arrow of thermodynamics.
• 🎯 Lost work = T₀·S_gen (Gouy–Stodola) — irreversibility directly destroys available work.
• 🎯 Entropy change of an ideal gas: Δs = c_v ln(T₂/T₁) + R ln(v₂/v₁) = c_p ln(T₂/T₁) − R ln(p₂/p₁) — pick the convenient form.
• ❌ Common mistake: using Celsius in Carnot efficiency or COP — these depend on the absolute (kelvin) temperature ratio, so °C gives a wrong result.

Expected exam pattern

A 1-mark Carnot-efficiency or COP NAT, and a 2-mark entropy-change or availability/irreversibility problem. The COP relations, ΔS ≥ 0 principle, and lost-work = T₀·S_gen are frequently tested.

Quick recap

Kelvin–Planck & Clausius → Carnot η = 1 − T_c/T_h. COP_ref = T_c/(T_h−T_c), COP_HP = T_h/(T_h−T_c) = COP_ref + 1. Clausius ∮δQ/T ≤ 0; entropy dS = (δQ/T)_rev, ΔS_isolated ≥ 0. Availability = max useful work; lost work = T₀·S_gen. Use kelvin.

Cover the answer, recall, then check. 12 cards on the second law for GATE ME.

Q1. State the Kelvin–Planck statement.
A1. No cyclic device can convert all absorbed heat into work; some heat must be rejected to a cold reservoir.

Q2. State the Clausius statement.
A2. Heat cannot flow spontaneously from a colder to a hotter body without external work input.

Q3. Carnot (maximum) efficiency between two reservoirs?
A3. η = 1 − T_c/T_h, with temperatures in kelvin.

Q4. Heat-engine efficiency in terms of heat quantities?
A4. η = W/Q_h = 1 − Q_c/Q_h.

Q5. COP of a refrigerator (Carnot)?
A5. COP_ref = Q_c/W = T_c/(T_h − T_c).

Q6. COP of a heat pump (Carnot), and its relation to the refrigerator?
A6. COP_HP = Q_h/W = T_h/(T_h − T_c) = COP_ref + 1.

Q7. State the Clausius inequality.
A7. ∮δQ/T ≤ 0; equal to zero for a reversible cycle, negative for an irreversible one.

Q8. Define entropy change.
A8. dS = (δQ/T)_rev — entropy is a property; its change is found via a reversible path.

Q9. State the increase-of-entropy principle.
A9. For an isolated system, ΔS ≥ 0; entropy stays constant only for a reversible process.

Q10. Entropy change of an ideal gas between two states?
A10. Δs = c_v·ln(T₂/T₁) + R·ln(v₂/v₁) = c_p·ln(T₂/T₁) − R·ln(p₂/p₁).

Q11. Define availability (exergy).
A11. The maximum useful work obtainable as a system reaches equilibrium with the surroundings (the dead state).

Q12. State the Gouy–Stodola relation.
A12. Lost (destroyed) work = T₀·S_gen — irreversibility, quantified by entropy generation, destroys available work.

The first law says energy is conserved; the second law says not all energy is equally useful and processes have a preferred direction. Entropy quantifies irreversibility, and availability (exergy) measures the maximum useful work. GATE tests Carnot limits, entropy change, and exergy destruction.

Core concept: the second law caps how much heat can become work (Carnot efficiency) and states that the entropy of an isolated system can only increase; availability is the useful-work potential relative to the environment.

Deep explanation

Beginner — second-law statements and Carnot

• Kelvin–Planck: no cyclic device can convert heat entirely into work (some must be rejected).
• Clausius: heat cannot flow from cold to hot without external work.
• Carnot efficiency (the maximum for any heat engine between T_H and T_C):
  η_Carnot = 1 − T_C/T_H (temperatures in kelvin).
  No real engine can beat it; a Carnot refrigerator has COP = T_C/(T_H − T_C).

Intermediate — entropy

Entropy change: dS = δQ_rev/T. For processes:

• Reversible adiabatic (isentropic): ΔS = 0.
• Ideal gas: ΔS = mc_v ln(T₂/T₁) + mR ln(V₂/V₁) = mc_p ln(T₂/T₁) − mR ln(p₂/p₁).
• Clausius inequality: ∮δQ/T ≤ 0; equality for reversible, "<" for irreversible cycles.
  Entropy generation S_gen ≥ 0 measures irreversibility (friction, heat transfer across finite ΔT, mixing).

Advanced — availability (exergy)

Availability is the maximum useful work obtainable as a system comes to equilibrium with the environment (at T₀). For a closed system: Φ = (U − U₀) + p₀(V − V₀) − T₀(S − S₀). For steady flow: ψ = (h − h₀) − T₀(s − s₀) + ke + pe.

• Gouy–Stodola theorem: lost work (irreversibility) I = T₀·S_gen — every bit of entropy generated destroys T₀·S_gen of useful work.
• Second-law (exergetic) efficiency compares actual work to the reversible maximum, exposing where a plant wastes potential even when the first law looks fine.

Worked example

A heat engine operates between 800 K and 300 K. Find the maximum possible efficiency, and the maximum work from 1000 kJ of heat input.
η_Carnot = 1 − T_C/T_H = 1 − 300/800 = 1 − 0.375 = 0.625 or 62.5%.
Maximum work = η × Q_in = 0.625 × 1000 = 625 kJ. No real engine between these reservoirs can exceed this; the remaining 375 kJ must be rejected to the 300 K sink.

GATE relevance

Carnot efficiency/COP, entropy change of ideal gases, Clausius inequality, and availability/exergy destruction (I = T₀S_gen) are core Thermodynamics questions. The T_C/T_H structure and the sign of entropy generation are frequently tested.

Exam tricks & shortcuts

• Carnot uses absolute (kelvin) temperatures — convert from °C first.
• Any claimed efficiency above η_Carnot = 1 − T_C/T_H is impossible (violates the second law) — a common conceptual question.
• Irreversibility (lost work) = T₀ × S_gen (Gouy–Stodola).
• Mnemonic: "One minus cold-over-hot, the best you've got."

Key formulas

• Carnot efficiency: η = 1 − T_L/T_H; COP (fridge) = T_L/(T_H − T_L).
• Entropy: dS = δQ_rev/T; Clausius inequality ∮δQ/T ≤ 0.
• Entropy generation: ΔS_universe ≥ 0.
• Availability (exergy): Φ = (u − u₀) + P₀(v − v₀) − T₀(s − s₀).
• Reversible work = maximum useful work.

The second law sets efficiency limits and the direction of processes; exergy quantifies the useful work potential.

Worked Example

Problem: A Carnot engine operates between a hot reservoir at 800 K and a cold reservoir at 300 K. It absorbs 1000 kJ from the hot reservoir. Find (a) the thermal efficiency, (b) the work output and heat rejected, and (c) the net entropy change of the two reservoirs.

Solution:
(a) Carnot (maximum) efficiency depends only on the reservoir temperatures:
η = 1 − T_C/T_H = 1 − 300/800 = 1 − 0.375 = 0.625 = 62.5%.

(b) Work output:
W = η·Q_H = 0.625 × 1000 = 625 kJ.
Heat rejected to the cold reservoir:
Q_C = Q_H − W = 1000 − 625 = 375 kJ.

(c) Entropy changes of the reservoirs:
Hot reservoir loses heat: ΔS_H = −Q_H/T_H = −1000/800 = −1.25 kJ/K.
Cold reservoir gains heat: ΔS_C = +Q_C/T_C = +375/300 = +1.25 kJ/K.
Net entropy change: ΔS_net = −1.25 + 1.25 = 0.

The zero net entropy change confirms the Carnot cycle is reversible.

Answer: (a) η = 62.5%; (b) W = 625 kJ, Q_C = 375 kJ; (c) ΔS_net = 0 (reversible).